What went wrong in these solutions of $\log \big(x^{\log x}\big)=4$

Question

Attempting to solve the following equation, I got multiple different answers. Can someone explain what I did wrong and show me how to solve the problem correctly using the same method? Thanks

$$\huge\log (x^{\log x})=4$$

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Note: $\log$ used in this question is $\log_{10}$

"Solution" 1:

Exponentiating, I get $$\large10^{\log x^{\log x}}=10^4$$ which simplifies to $$\large \log x=10000$$Solving for $x$, I get the answer of $$x=10^{10000}\quad\text{(obviously wrong)}$$

"Solution" 2:

Using the logarithmetic power rule, I simplify this equation to $$\large(\log x)^2=4$$ which simplifies to $$\large\log x=\pm 2$$ Solving these two equations, I get that $$x=100\;\text{or}\;x=\dfrac1{100}$$

Wolfram and Symbolab agree with the second solution, while Mathway gives a result of $x=10000$.

Answer 1

EDIT : I assume that you mean $\log(x^{\log(x)})$.

Whenever you take the logarithm of something which is of the form $x^a$ then

$$\log(\lambda^a) \equiv a \log(\lambda).\,\,\,(♦)$$

What I mean to say is that the exponent "comes down".

Taking $\log(x^{\log(x)})$ and comparing it with $(♦)$, we get $\lambda = x$ and $a = \log(x)$. So the next steps should be clear :

$$\log(x^{\log(x)}) = (\log(x))^2 =4$$

$$\implies \log(x) = \pm 2$$

$$x = 10^2 \,\,\,\,\,\,\,OR \,\,\,\,\,x=10^{-2} = \dfrac{1}{100}$$

So your "second" solution is correct except for the first step where you wrote $LHS=2$ instead of $4$.

($LHS\,\,\equiv $ "Left Hand Side" (of the equation) )

Now coming to your first solution :

Note the following property :

$$b^{\log_b(\lambda)} \equiv \lambda\,\,\,\,.(♣)$$

That is, if both the "number to be raised to some power" and "base of logarithm in the exponent" are same, then what we get is "the number/quantity whose logarithm is being taken in the exponent".

Come to the original equation : $\log_{10}(x^{\log(x)}) =4$.

Now raise both sides of the equation to $10$ :

$$10^{\log_{10}(x^{\log(x)})} = 10^4$$

Compare this with $(♣)$ to get : $b=10$ and $\lambda = x^{\log(x)}$. So what we should get is :

$$x^{\log(x)} = 10000$$

That seems to bring us nowhere. Though it is possible to continue from here, the main point I want to highlight is the mistake you have committed in your "first" solution.

I have tried to make it as detailed as possible.

Hope this helps ! :-)

Answer 2

In the solution 1, how do you get the second equality? In fact, $$10^{\log x^{\log x}}=x^{\log x},$$ and the equality becomes $$x^{\log x}=10^4 \Leftrightarrow (\log x)^2=4.$$